Question
Calculate the mass of silver produced when a current of 0.5A is passed through AgNO3 for 1hour 30mins.
Answers
DrBob222
coulombs =amperes x seconds
C = 0.5 ampere x 1.5 hrs x (60 min/hr) x (60 sec/min) = 5400
96,485 coulombs will deposit 107.9 g Ag and you have 5400 so
107.9 g Ag x (5.400/96,485) = ?
C = 0.5 ampere x 1.5 hrs x (60 min/hr) x (60 sec/min) = 5400
96,485 coulombs will deposit 107.9 g Ag and you have 5400 so
107.9 g Ag x (5.400/96,485) = ?